Question

For tin the conductivity tensor is diagonal, with entries \(a, a\), and \(b\) when referred to its crystal axes. A single crystal is grown in the shape of a long wire of length \(L\) and radius \(r\), the axis of the wire making polar angle \(\theta\) with respect to the crystal's 3-axis. Show that the resistance of the wire is \(L\left(\pi r^{2} a b\right)^{-1}\left(a \cos ^{2} \theta+b \sin ^{2} \theta\right)\).

Short answer

The resistance of the wire is \( L ( \pi r^2 a b )^{-1} ( a \cos^2 \theta + b \sin^2 \theta ) \).

Step-by-step solution

- Understand the problem

We need to determine the resistance of a wire made from tin whose conductivity tensor is diagonal with entries \(a, a, b\) in the crystal axes. The wire's axis makes an angle \(\theta\) with the crystal's 3-axis.

- Write down the conductivity tensor

The conductivity tensor \(\sigma \) in the crystal axes is given by: \[ \sigma = \begin{pmatrix} a & 0 & 0 \ 0 & a & 0 \ 0 & 0 & b \end{pmatrix} \]

- Transform the tensor to the wire's coordinate system

Use the rotation matrix \( R \), which rotates the coordinate system by angle \( \theta \): \[ R = \begin{pmatrix} \cos \theta & 0 & \sin \theta \ 0 & 1 & 0 \ -\sin \theta & 0 & \cos \theta \end{pmatrix} \] The new conductivity tensor \( \sigma' \) in the wire's coordinates is \( \sigma' = R^T \sigma R \).

- Calculate the transformed tensor

Perform the matrix multiplication to find \( \sigma' \): \[ \sigma' = \begin{pmatrix} \cos \theta & 0 & -\sin \theta \ 0 & 1 & 0 \ \sin \theta & 0 & \cos \theta \end{pmatrix} \begin{pmatrix} a & 0 & 0 \ 0 & a & 0 \ 0 & 0 & b \end{pmatrix} \begin{pmatrix} \cos \theta & 0 & \sin \theta \ 0 & 1 & 0 \ -\sin \theta & 0 & \cos \theta \end{pmatrix} = \begin{pmatrix} a \cos^2 \theta + b \sin^2 \theta & 0 & (a-b) \sin \theta \cos \theta \ 0 & a & 0 \ (a-b) \sin \theta \cos \theta & 0 & a \sin^2 \theta + b \cos^2 \theta \end{pmatrix} \]

- Use the appropriate component for resistance computation

The relevant component for the conductivity along the wire's axis is \( \sigma_{11}' = a \cos^2 \theta + b \sin^2 \theta \).

- Express resistance in terms of conductivity

Recall the relationship between resistance \(R\), length \(L\), cross-sectional area \(A = \pi r^2\), and conductivity \( \sigma \: R = \frac{L}{A \sigma} \). Substituting the values, we get \ R = \frac{L}{\pi r^2 (a \cos^2 \theta + b \sin^2 \theta)} \

Key concepts

resistance computation

The resistance of a material is an important electrical property that defines how much it opposes the flow of electric current. For the given wire made of tin, we need to compute the resistance based on its physical dimensions and the conductivity tensor. The wire's length is denoted by \(L\), and the radius is \(r\). Thus, the cross-sectional area \(A\) of the wire is \(A = \pi r^2\). To connect the resistance \(R\) and conductivity \(\sigma\), we use the formula:

\[ R = \frac{L}{A \sigma} \]
According to this, the resistance is inversely proportional to both the conductivity of the wire’s material and the cross-sectional area, and directly proportional to its length. For our specific problem, the transformed component of the conductivity tensor that aligns with the wire's axis, denoted as \(\sigma_{11}'\), is used in the formula. This means:
\[ R = \frac{L}{\pi r^2 (a \cos^2 \theta + b \sin^2 \theta)} \]

tensor transformation

A conductivity tensor provides a compact representation of how conductive a material is in different directions. In the given problem, the initial conductivity tensor for tin in its crystal axis is diagonal, meaning that conductivity values are given by:\[ \sigma = \begin{pmatrix} a & 0 & 0 \ 0 & a & 0 \ 0 & 0 & b \end{pmatrix} \]

To determine the conductivity along the axis of the wire which makes an angle \(\theta\) with the crystal's 3-axis, we need to transform this tensor to align it with the wire's coordinate system. This is achieved through a rotation matrix \(R\):
\[ R = \begin{pmatrix} \cos \theta & 0 & \sin \theta \ 0 & 1 & 0 \ -\sin \theta & 0 & \cos \theta \end{pmatrix} \]
By rotating the original tensor using the relation \(\sigma' = R^T \sigma R\), we obtain the transformed tensor:\[ \sigma' = \begin{pmatrix} a \cos^2 \theta + b \sin^2 \theta & 0 & (a-b) \sin \theta \cos \theta \ 0 & a & 0 \ (a-b) \sin \theta \cos \theta & 0 & a \sin^2 \theta + b \cos^2 \theta \end{pmatrix} \] The relevant component here for our resistance calculation is \(\sigma_{11}' = a \cos^2 \theta + b \sin^2 \theta\).

crystal axes

In materials science, the crystal axes are reference directions that align with the symmetrical properties of the crystal lattice. For the given tin crystal, the conductivity tensor is initially specified along these crystal axes, which gives us insights into the anisotropic (direction-dependent) electrical properties of the material. In our exercise, the conductivity tensor is diagonal with entries \(a, a\), and \(b\), signifying different conductivities along different crystallographic directions.

An important aspect of this problem is determining the effective conductivity of the crystal when the wire's axis (length \(L\)) makes a polar angle \(\theta\) with the crystal's 3-axis. The polar angle specifies how rotated the wire is relative to one of the crystal's principal axes and fundamentally affects how we perceive electrical conductivity along the wire's axis. Thus, understanding the crystal axes and how to transform the tensor from these axes to the wire's coordinate system is crucial for accurately calculating the resistance.